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材料科学与工程基础作业讲评-8_图文

第十二次作业 中文
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4-41 在20 ℃时1m3内含有1021个电荷载体的半导
体的电阻率为 0.1Ω·m ,如果在电场为 0.15V· mm-1 时可传导1A的电流,试计算电导的平均漂移速度

? ?

?=nq?=nqv/E, so, v=?E/nq= E/(nq?)

?

=150/[1021*1.6*10-19*0.1]=9.375m/s
场中的迁移速度。

? 载流子的迁移率,其物理意义为载流子在单位电

4-43已知硅和锗在300 ℃的电阻率分别为 2.3×103μΩ· m和0.46μΩ· m,试分别计算硅和锗在 250℃的电导率。
4-43: ln? =C -[ E g /(2kT)], 或?=?0exp[-Eg/(2kT)] ln(?1/ ?2 )= -Eg/(2k)[1/T1-1/T2] , R=1/? 硅半导体 (Eg=1.2ev, k=8.62?10-5ev): 250?C, ? =0.15?103(Ωm)-1 锗半导体 (Eg=0.67ev, k=8.62?10-5ev): 250?C, ? =1.14?106(Ωm)-1

4-49 产生介电损耗的原因是什么?主要的影响因 素有哪些?
定义:电介质在交变电场作用下,电能转变成热能 而损耗。或P357。 主要原因:漏电电流和极化电流损耗 主要影响因素: (1) 分子结构:极性大,tg? 大 ;基团数目 多 tg? 大 ; (2) 小分子及杂质;(3) 多相体系; (4) 交变电场频率; (5) 温度

英文书:
12.11 (a) Calculate the drift velocity of electrons

in germanium at room temperature and when the magnitude of the electric field is 1000 V/m. (b) Under these circumstances, how long does it take an electron to traverse a 25 mm (1 in.) length of crystal? Solution: (a) v =? E=0.38m2.(Vs-1)?1000V/m =380m/s (b) 25mm/(380m/s)= 6.6? 10-5s

12.14:(a) Calculate the number of free electrons per cubic meter for gold assuming that there are 1.5 free electrons per gold atom. The electrical conductivity and density for Au are 4.3?107(Ω-m)-1 and 19.32 g/cm3, respectively. (b) Now compute the electron mobility for Au. Solution: n=1.5?(?/M)?NA= 1.5?[19.32?106(g/m3) ?197(g/mol)]?6.02?1023=8.86?1028 m-3 ? =n|e|?e So, ?e=?/[n|e|] =(4.3 ?107Ω-1m-1) ? [(8.86?1028 m-3)(1.6?10-19C)] =3.03?10-3 m2/V-s

12.38 The intrinsic electrical conductivities of a semiconductor at 20 and 100?C are 1.0 and 500 (Ω-m)-1, respectively. Determine the approximate band gap energy for this material. Solution: ln?=C-Eg/(2kT) so, ln?1=C-Eg/(2kT1), ln?2=C-Eg/(2kT2) ln?1- ln?2=Eg/(2kT2 )-Eg/(2kT1)=(Eg/2k)[1/T2 -1/T1)] so, Eg= (ln?1- ln?2)?2k ?[1/T2-1/T1] = (ln1- ln500)?2?8.62?10-5eV/K ?(1/373 -1/293K) =1.46eV

12.51 At temperatures between 775?C and 1100?C, the activation energy and pre-exponential for the diffusion coefficient of Fe2+ in FeO are 102,000 J/mol and 7.3?10-8m2/s, respectively. Compute the mobility for an Fe2+ ion at 1000?C.

Solution:
D=D0exp(-Qd/RT) =7.3×10-8exp[-102000/(8.314×1273)] =4.76×10-11m2/s ? =n e D/(k T) P390 =2×1.6×10-19C× 4.76×10-11m2/s ?(1.381?10-23J/K×1273K)= 8.7×10-10 m2/V-s ? =n e D/(k T) P390 =2×4.76×10-11m2/s ?(8.62?10-5ev/K×1273K)=8.7×10-10 m2/V-s

思考题
4-39试述影响离子电导率和电子电导率的主要影响因素 离子电导率: 温度:温度越高,电导率越大 晶体结构:离子大小、电荷影响电导率 晶格缺陷:离子性晶格缺陷的生成及浓度大小是 决定离子电导的关键 电子电导率(金属):散射

温度 σ ∝T-1 杂质及缺陷 杂质原子(离子)引入新的局部能级 塑性形变 位错

4-47.在下列高聚物材料中,哪些有 可能利用高频塑化法加工成型?(1) 酚醛树脂;(2)聚乙烯;(3)聚 苯乙烯;(4)聚氯乙烯
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4-47 聚氯乙烯

12.8 In terms of electron energy band structure, discuss reasons for the difference in electrical conductivity between metals, semiconductors, and insulators.

12.9 If a metallic material is cooled through its melting temperature at an extremely rapid rate, it will form a noncrystalline solid (i.e., a metallic glass). Will the electrical conductivity of the noncrystalline metal be greater or less than its crystalline counterpart? Why? 电导率降低。英文书P374,和P375-图12.8
(温度升高,金属电阻率增加,电导率下降,导电性减弱)

12.21 (a) Compute the number of free electrons and holes that exist in intrinsic germanium at room temperature, using the data in Table 12.2. (b) Now calculate the number of free electrons per atom for germanium and silicon (Example Problem 12.1). (c) Explain the difference. You will need the densities for Ge and Si, which are 5.32 and 2.33 g/cm3, respectively. Solution: (a) n=p= ?/[|e|(?e+?h)] =2.2( -m)-1?[(1.6?10-19C)(0.38+0.18 m2/V-s)] = 2.46?1019 m-3

(b) for Ge: Number of atoms: N= (?/M)?NA = [5.32?106(g/m3) ?72.59(g/mol)]?6.02?1023 =4.41?1028 m-3 每个原子自由电子数=n/N=(2.46?1019 m-3 )/(4.41?1028 m-3 ) =5.6?10-10 for Si: Number of atoms: N= (?/M)?NA = [2.23?106(g/m3) ?28.1(g/mol)]?6.02?1023 =4.78?1028 m-3 n=p= ?/[|e|(?e+?h)] =4?10-4( -m)-1?[(1.6?10-19C)(0.14+0.048 m2/V-s)] =1.33?1016 m-3 每个原子自由电子数=n/N=(1.33?1016 m-3 )/(4.78?1028 m-3 ) =2.78?10-13 (c) 每个原子的自由电子数差别大,电导率差别也大。

12.34 Compare the temperature dependence of the conductivity for metals and intrinsic semiconductors. Briefly explain the difference in behavior.
金属减小;半导体为增大。

第十三次 中文
4-52、 铁磁性材料和 顺磁性材料的 磁性来 源有何异同?
铁磁性来源于很强的永久磁矩, 内部交换场 顺磁性来源于永久磁矩

4-55、将2400A.m-1的磁场作用到相对磁导率 为5000的材料上,计算磁感强度和磁化强度。

4-55:磁化强度: M=(μr-1)H=(5000-1)?2400A.m-1 =1.2?107A.m-1 磁感强度: B= μ0 (H+M)=4??10-7(H.m-1) ?(2400A.m-1 +1.2?107A.m-1) =15.1Wb.m-2=15.1tesla B= μr μ0H=5000× 4π×10-7H/m ×2400A/m=15.1tesla

18.7 The magnetization within a bar of some metal alloy is 3.2 ?105 A/m at an H field of 50 A/m. Compute the following: (a) the magnetic susceptibility, (b) the permeability, and (c) the magnetic flux density within this material. (d) What type(s) of magnetism would you suggest as being displayed by this material? Why?

(a)χm=M/H=(3.2×105A/m)/(50A/m)=6400 (b)μr=χm+1=6401 μ=μrμ0=2.56π×10-3H/m ( c ) B=μH=2.56π×10-3H/m× ( 50A/m ) =0.402tesla (d)铁磁体

18.8 Compute (a) the saturation magnetization and (b) the saturation flux density for cobalt, which has a net magnetic moment per atom of 1.72 Bohr magnetons and a density of 8.90 g/cm3.
a)Ms=1.72μBN=1.72μBρNA/ACo

=1.72×9.27×10-24×8.9×106×6.023×1023/58.9 =1.45×106A/m
(b)Bs=μ0Ms=(4π×10-7H/m)· Ms=1.822tesla

18.9 Confirm that there are 2.2 Bohr magnetons associated with each iron atom, given that the saturation magnetization is 1.70× 106 A/m, that iron has a BCC crystal structure, and that the unit cell edge length is 0.2866 nm.

将number of Bohr magnetons 代入公式计算Ms Ms=nBμB/a3 =2×2.2×9.27×10-24/(0.2866×10-9m)3 =1.733×106A/m

18.25 Figure 18.25 shows the B-versus-H curve for a steel alloy (a) What is the saturation flux density? (b) What is the saturation magnetization? (c) What is the remanence? (d) What is the coercivity? (e) On the basis of the data in Tables 18.5 and 18.6, would you classify this material as a soft or hard magnetic material? Why?

remanence

saturation flux density

coercivity

软磁性

Solution
(a) The saturation flux density
(b) Bs= ?0(Hs+Ms)? ?0Ms Bs=1.4 tesla

Ms=Bs/ μ0=1.4/(4π×10-7H/m)=1.11 ×106A/m
(c) What is the remanence? Br=0.8 tesla (d) What is the coercivity? Hc=75 A/m (e) On the basis of the data in Tables 18.5 and 18.6, would you classify this material as a soft or hard magnetic material? Why?

思考题:
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4-50 写出下列物理量的量纲:( 1 )磁感应 强度;( 2 )磁化强度;( 3 )磁导率和相 对磁导率;(4)磁化率。

4-54试述影响金属磁性的因素。 ? 正离子的抗磁性和顺磁性; 电子的抗磁性和顺磁性。 四个主要因素。
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18.5 (a) Explain the two sources of magnetic moments for electrons. 电子的轨道磁矩和 自旋磁矩。 (b) Do all electrons have a net magnetic moment? Why or why not? 是。 (c) Do all atoms have a net magnetic moment? Why or why no?非,具有各层都 充满电子的原子结构,其电子磁矩相互抵 消,不显磁性。

18.12 Cite the major similarities and differences between ferromagnetic and ferrimagnetic materials. ? Ferromagnetic-铁磁性 ? Ferri-magnetic-亚铁磁性-保留了剩 余磁矩。磁化率比铁磁性低。
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18.19 Briefly explain why the magnitude of the saturation magnetization decreases with increasing temperature for ferromagnetic materials, and why ferromagnetic behavior ceases above the Curie temperature. For ferromagnetic, the atomic thermal motions counteract the coupling forces between the adjacent atomic dipole moments, causing some dipole misalignment, regardless of whether an external field is present. This results in a decrease in the saturation magnetization for both ferro- and ferrimagnets. The saturation magnetization is a maximum at 0 K, at which temperature the thermal vibrations are a minimum. With increasing temperature, the saturation magnetization diminishes gradually and then abruptly drops to zero at what is called the Curie temperature Tc . Section 18.6


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